matriks-08

8/8/2019 Matriks-08

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Garis-garis Besar Perkuliahan

1. Sistem Persamaan Linier

2. Matriks dan Operasinya

3. Determinan dan Sifat-sifatnya

UTS

4. Ruang Vektor

5. Basis dan Dimensi

6. Ruang Hasil Kali Dalam

7. Transformasi Linier

UTS

8. Matriks Transformasi Linier

9. Keserupaan

10. Nilai Eigen dan Vektor Eigen

11. Diagonalisasi

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MATRICES

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Introductory

Two matrices are equal if they are of the same size and if their

corresponding elements are equal.

a ij : the element of matrix A in row i and column j .

For a square n vn matrix A, the main diagonal is:

11 12 1

21 22 2

1 2

n

n

n n nn

a a a

a a a

a a

A

a

« »

¬ ¼¬ ¼!¬ ¼¬ ¼- ½

L

L

M M O M

L

Thus A = B if a ij

= b ij .

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Addition

Let A and B be matrices of the same size.

Their sum A + B is the matrix obtained by adding together the

corresponding elements of A and B.

The matrix A + B will be of the same size as A and B.

If A and B are not of the same size, they cannot be added, and we

say that the sum does not exist.

ijijij bac B AC !! thenif Thus ,

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Examples 1 and 2

1 2

2,

1 2,Let and A B C

« » « » « »! ! !¬ ¼ ¬ ¼ ¬ ¼- ½ - -

½ ½

Determine A + B and A + C , if the sum exist.

Solution

1 2 4 5 7 6

0 3 2 1 3

1 4 7 2 5 6

0 2 3 3 1

8

3 9 1

3 1 1

( )

1

1

.

8 A B

« » « » ! ¬ ¼ ¬ ¼- ½ - ½

« »

! ¬ ¼- ½

« »! ¬

-

¼½

(2) Because A is 2 v 3 matrix and C is a 2 v 2 matrix, they are not

of the same size, A + C does not exist.

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Scalar Multiplication

Let A be a matrix and c be a scalar. The scalar multiple of A by c ,

denoted cA, is the matrix obtained by multiplying every element of

A by c . The matrix cA will be the same size as A.

Example 3

1 2 4

7 2 0.

Let A

« »!

¬- ¼½

3 1 3 ( 2) 3 4

3 7 3 ( 3) 33 .

3 6 1

10

2

2 9 0 A

v v v

v v v

« » « »! !¬ ¼ ¬ ¼

- ½ - ½

Observe that A and 3 A are both 2 v 3 matrices.

ijij cabc !! theniThus ,

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Negation and Subtraction

We now define subtraction of matrices in such a way that makes it

compatible with addition, scalar multiplication, and negative. Let

A ² B = A + ( ²1 )B

Example 4

5 0 2 2 8 13 6 5 .0 4 6« » « »! ¬ ¼ ¬ ¼- ½ ½ - ASu se a d

5 2 0 8 2 ( 1) 3 8 1

3 0 6 4

.

5 6 3 2 1 1

« » « » ! !¬ ¼ ¬ ¼

- ½ -

½

A B

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Multiplication

Let the number of columns in a matrix A be the same as thenumber of rows in a matrix B. The product AB then exists.

If the number of columns in A does not equal the number of rows

in B, we say that the product does not exist.

Let A: m vn matrix, B: n vk matrix,

The product matrix C = AB has elements

C is a m vk matrix.

11 1 2 2

1

2

2

i iij i j i j i in

j

j

n

j

nj

n

a a a

b

c a b a b bb

b

a

« »¬ ¼¬ ¼

! «! »- ½ ¬ ¼¬ ¼¬ ¼- ½

LM

L

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Example 5

? A6 2

5 0 1

3 2 6

1 3

2 0 5 .Le t a d

Deter i e a d i f t e r duc ts exi st .

A B C

A B B A A C

« » « »

! ! !¬ ¼ ¬ ¼- ½- ½

51 3

2

0

60

1

3 2

A B« » « »

! ¬ ¼ ¬ ¼

- ½ - ½

(1 5) (3 3) (1 0) (3 ( 2)) (1 1) (3 6)

(2 5) ( 3) (2 0) (0 ( 2)) (2 1) (0 6)

14 6 1

10.

0 2

0

9

« v v v v v v

v v v v

»! ¬ ¼

- ½

« »! ¬ ¼

-

v

½

v

? A ? A ? A

? A ? A ? A

1 3 1 3 1 3

2 0 2 0 2

5 0 1

3 2 6

5 0 10

3 2 6

« » « » « »¬ ¼ ¬ ¼ ¬ ¼

- ½ - ½ - ½

« » « » « »¬ ¼ ¬ ¼ ¬ ¼

-

« »¬ ¼¬ ¼! ¬ ¼¬ ¼¬ ¼- ½½ - ½ - ½

BA and AC do not exist.

Solution

Note: In general, AB { BA.

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? A2)14()23(1

2

43!vv!

¼½

»

¬-

«

Example 6

.5301

and2307

12

et ¼½»

¬-«

!¼¼½

»

¬¬-

«

! B A

? A ? A

? A ? A

? A ? A ¼¼

¼¼¼

½

»

¬¬

¬¬¬

-

«

¼½»

¬-«

¼½»

¬-«

¼½

»¬-

«¼½

»¬-

«¼½»

¬-«

¼½»

¬-«

!¼½

»¬-

«

¼

¼

½

»

¬

¬

-

«

!

50

2331

23

50

0731

07

50

1231

12

5301

23

0712

AB

¼¼

½

»

¬¬

-

«

!

¼¼

½

»

¬¬

-

«

!

1030751

1006300075032

!23c

Determine AB.

¼½»

¬-«

!¼½»

¬-«

!

105237

and4312

B A

Example 7

Let C = AB Determine c 23.

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A zero matrix is a matrix in which all the elements are zeros.

A diagonal matrix is a square matrix in which all the elements not

on the main diagonal are zeros.

An identity matrix is a diagonal matrix in which every diagonal

element is 1.

0 0 0

0 0 0

0 0 0

zero matrix

mn0

« »¬ ¼¬ ¼

! ¬ ¼¬ ¼- ½

11

22

0 0

0 0

0 0

diagonal matrix A

L

L

M M L M

Lnn

a

a

a

A

« »¬ ¼¬ ¼

! ¬ ¼¬ ¼- ½

1 0 0

0 1 0

0 0 1

identity matrix

L

L

M M L M

L

n I

« »¬ ¼

¬ ¼! ¬ ¼¬ ¼- ½

Special Matrices

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Some PropertiesLet A be m v n matrix and 0mn be the zero m v n matrix. Let B be an

n v n square matrix. 0n and I n be the zero and identity n v n matrices. Then

A + 0mn = 0mn + A = A

B0n = 0n B = 0n

BI n = I n B = B

Example 9

.4312

and854312

et¼½»

¬-«

!¼½»

¬-« ! B A

23

2 1 3 0 0 0 2 1 3

4 5 8 0 0 0 4 5 8 A A

« » « » « » ! ! !¬ ¼ ¬ ¼ ¬ ¼- ½ - ½ - ½0

2 2

2 1 0 0 0 0

3 3 0 0 0 0B

« » « » « »! ! !¬ ¼ ¬ ¼ ¬ ¼

- ½ - ½ - ½0 0

2

2 1 1 0 2 1

3 4 0 1 3 4B I B

« » « » « »! ! !

¬ ¼ ¬ ¼ ¬ ¼ - ½ - ½ - ½

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Algebraic Properties

Let A, B, and C be matrices and a , b , and c be scalars. Assume

that the sizes of the matrices are such that the operations can

be performed.

Properties of Matrix Addition and scalar Multiplication

1. A + B = B + A Commutative p r ope r ty of

addition

2. A + ( B + C ) = ( A + B ) + C Associative pr ope r ty of addition

3. A + 0 = 0 + A = A ( whe r e 0 is the appr opr iate ze r o mat r ix )

4. c ( A + B ) = cA + cB Dist r ibutive pr ope r ty of addition

5. ( a + b )C = aC + bC Dist r ibutive pr ope r ty of addition

6. ( ab )C = a ( bC )

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Let A, B, and C be matrices and a , b , and c be scalars. Assume

that the sizes of the matrices are such that the operations can

be performed.

P

roperties

of Matrix Mul

tipli

cation

1. A( BC ) = ( AB )C Associative pr ope r ty of multiplication

2. A( B + C ) = AB + AC Dist r ibutive pr ope r ty of multiplication

3. ( A + B )C = AC + BC Dist r ibutive pr ope r ty of multiplication

4. AI n = I n A = A ( whe r e I n is the appr opr iate identity mat r ix )

5. c ( AB ) = ( cA )B = A( cB )

Note: AB { BA in general. M ultiplication of mat r ices is not

commutative .

Algebraic Properties

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¼½»

¬-«

¼½»

¬-«

¼½»

¬-«

1520

1873

5431

.5964

115584273031

¼½»

¬-« !

¼½»

¬-«

!

! C B A

.1520

and1873

5431

et¼½»

¬-«

!

¼½»

¬-« !

¼½»

¬-«

! C B A

Example 10

A+B=B+A

.)()( ijijijijijij A Babba B A !!!

Consider the ( i , j )th elements of matrices A+B and B+ A:

@ A+ B= B+ A

Proof

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Example 11

.1131

112

201

310

13

21

¼½

»

¬-

« !

¼½

»

¬-

«

¼½

»

¬-

«

! AB

.19

014

1131112

)(¼½»

¬-«

!

¼¼½

»

¬¬-

«

¼½»

¬-«

!C AB

.01

4

and201

310

13

21

Let ¼¼½

»

¬¬-

«

!

¼½

»

¬-

«

!

¼½

»

¬-

«

!

C B A Compute ABC .

Solution

( 1 ) ( AB )C

(2) A( BC )

¼½»

¬-«!

¼¼½

»

¬¬-

«

¼½»

¬-«

!

41

014

201310

BC

.1

9

4

1

13

21)(

¼½

»

¬-

«!

¼½

»

¬-

«

¼½

»

¬-

«

! BC A

Count the number of multiplications. Which method is better?

2v6+3v2=12+6=18

3v2+2v2=6+4=10

@ A( BC ) is better.

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In algebra we know that the following cancellation laws apply.

If ab = ac and a { 0 then b = c .

If pq = 0 then p = 0 or q = 0.

However the corresponding results are not true for matrices.

AB = AC does not imply that B = C .

PQ = O does not imply that P = O or Q = O.

Caution

Example 12

. but86

43thatObserve

.23

83 and

12

21

42

21 matricesheConsider t(1)

C B AC AB

C B A

{¼½

»¬-

«!!

¼½

»¬-

«

!¼

½

»¬-

«!¼

½

»¬-

«!

.and but,thatObserve

.31

62 and,

42

21 matricesheConsider t(2)

OQO P O PQ

Q P

{{!

¼½

»¬-

«

!¼

½

»¬-

«

!

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Powers

If A is an n v n square matrix and r and s are nonnegative integers,

then1. Ar As = Ar +s .

2. ( Ar )s = Ar s .

3. A0 = I n ( by definition )

Definition

If A is a square matrix, then

.timesk

k A AA A !

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Example 13

.01

214

A A computeIf ¼½

»

¬-

«

!

¼½

»¬-

«

!¼

½

»¬-

«

¼½

»¬-

«

!

21

23

01

21

01

212

A

.65

1011

21

23

21

234

¼½

»¬-

«

!¼

½

»¬-

«

¼½

»¬-

«

! A

Solution

2

2222

22

463

57362

57)2(3)2(

BBA AB

ABB ABBA AB A

ABB AB ABB A A

!

!

Example 14 Simplify the following matrix expression.

ABB AB ABB A A 57)2(3)2( 22 Solution

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Systems of Linear Equations

A system of m linear equations in n variables is as follows.

11 1 1

1 1

1

n n

m mn n m

a x a x

a

b

b x a x

!

!

M M M M

L

LLet

11 1 1

1

1

and, ,

n

m mn n m

a a x

a a x

b

A B

b

X

« » « » « »¬ ¼ ¬ ¼ ¬ ¼

! ! !

¬ ¼ ¬ ¼ ¬ ¼¬ ¼ ¬ ¼ ¬ ¼- ½ - ½ - ½

L

M L M M M

L

We can write the system of equations in the matrix form

AX = B

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Transpose of a Matrix

The trans pose of a matrix A, denoted At , is the matrix whose

columns are the rows of the given matrix A.

i.e. ( ): : ,t t

ij ji

A A An n m m A !v v

Example 15

? A.431and654721

0872

!¼½»

¬-«

!¼½»

¬-«

! C B A

¼½»

¬-« !

0782t

A

¼¼½

»

¬¬-

«

!

675241

t B .431

¼¼½

»

¬¬-

«!

t C

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Properties of Transpose

Let A and B be matrices and c be a scalar. Assume that the

sizes of the matrices are such that the operations can be

performed.1. ( A + B )t = At + Bt T r anspose of a sum

2. ( cA )t = cAt T r anspose of a scala r multiple

3

. ( AB )

t

= B

t

A

t

T r anspose of a pr oduct 4. ( At )t = A

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? A

ni jni ji j

ni

i

i

jn j j jiijt

bababab

b

b

aaa AB AB

!

¼¼¼¼

½

»

¬¬¬¬

-

«

!!

.

/.

2211

2

1

21

)()(

Proof

( AB)t = Bt At

? A ni jni ji j

jn

j

j

niii

t t t t

ij

t t

bababa

a

a

a

bbb

A j Bi A j Bi A B

!

¼¼¼¼¼

½

»

¬¬¬¬¬

-

«

!

!!

./

. 2211

2

1

21

]o [row]ocolumn[]o[column]o row[)(

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Symmetric Matrix

¼¼¼

½

»

¬¬¬

-

«

¼¼½

»

¬¬-

«

¼½»¬-

« 639432329370

4201

384871 410

4552

match

match

A symmetric matrix is a matrix that is equal to its transpose.

jiaa A A jiij

t , i.e., , !!

Example 16

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Let A be a square matrix. The trace of A, denoted t r ( A ) is the sum

of the diagonal elements of A. Thus if A is an n v n matrix,

t r ( A ) = a 11 + a 22 + « + a nn

Example 17

Determine the trace of the matrix .037652214

¼¼½

»

¬¬-

«

! A

Solution

We get.10)5(4)( !! At r

Trace of a Matrix

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Properties of Trace

Let A and B be matrices and c be a scalar. Assume that the sizes of the matrices are such that the operations can be performed.

1. t r ( A + B ) = t r ( A ) + t r ( B )

2. t r ( AB ) = t r ( BA )

3. t r ( cA ) = c t r ( A )4. t r ( At ) = t r ( A )

Since the diagonal element of A + B are ( a 11

+b 11 ), ( a

22+b

22 ), «,

( a nn +b nn ), we get

t r ( A + B ) = ( a 11 + b 11 ) + ( a 22 + b 22 ) + «+ ( a nn + b nn )

= ( a 11 + a 22 + « + a nn ) + ( b 11 + b 22 + « + b nn )

= tr( A ) + tr( B ).

Pr oof of (1)

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The inverse of an invertible matrix is unique.

Pr oof

Let B and C be inverse of A. Thus AB = BA = I n , and AC = CA = I n .

Multiply both sides of the equation AB = I n by C .

C ( AB ) = CI n

( CA )B = C I n B = C

B = C

Thus an invertible matrix has only one inverse.

The Inverse of a Matrix

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How to find A-1?

? A ? A., 21211

nnnC C C I X X X A .. !! Let

We shall find A-1 by finding X 1, X 2, «, X .

Since AA

-1

=I n , then? A ? A

.2121 nnC C C X X X A

..!

.,,, 2211 nnC AX C AX C AX !!! .i.e.,

Solve these systems by using Gauss-Jordan elimination:

? A? A.:

:

21

21

nn

n

X X X I

C C C A

...

}} :matrixaug me ted

? A ? A.:: 1}}@ AI I A

nn.

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Example 19

Determine the inverse of the matrix¼¼½»

¬¬-«

!531532 211 A

Solution

¼¼½

»

¬¬-

«

!100531 010532

001211

]:[ 3 I A ¼¼½

»

¬¬-

«

}

101320 012110

001211

1132)(2

¼¼

½

»

¬¬

-

«

}

101320

012110001211

2)1(

¼

¼

½

»

¬

¬

-

«

}

123100

012110013101

2)2(3

21

¼¼½

»

¬¬-

«

}

123100135010010001

3)1(231 .

123135110

Thus, 1

¼¼½

»

¬¬-

«

!

A

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¼¼

½

»

¬¬

-

«

}

135000011210012301

3 23

2)1(1

Example 20

Determine the inverse of the following matrix, if it exists.

¼¼½

»

¬¬-

«

412

721

511

A

Solution

¼¼½

»

¬¬-

«

!

100412010721 001511]:[ 3 I A

¼¼½

»

¬¬-

«

}

102630011210 001511

1)2(3

1) 1(2

There is no need to proceed further.

The reduced echelon form cannot have a one in the ( 3, 3 ) location.

The reduced echelon form cannot be of the form [ I n : B ].

Thus, A ² 1 does not exist.

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Properties of Matrix Inverse

Let A and B be invertible matrices and c a nonzero scalar, Then

A A ! 11)( 1.

11 1)( .2 ! A

cc A

111)( .3 ! A B AB

nn A A )()( .4 11 !

t t A A )()( .5 11 !

Pr oof

1. By definition, AA-1= A-1 A=I .

))(())(( .2 11 11cA A I AcA

cc

!!

))(()())(( .31111111

AB A B I AA A BB A A B AB

!!!!

nn

k k

nn A A I A A A A A A )()( .41

times

11

times

1 !!! ..

,)( )( ,

,)( )( , .5

111

111

I A A A A I A A

I A A AA I AA

t t t

t t t

!!!

!!!

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.)

.43

11,

13

14

1

1

¼½

»¬-

«

!¼

½

»¬-

«!

t A

A A

( computetoninformatio

thisUsethatshownbecanitthenIf

Solution

.)()(4131

431111

¼½

»¬-

«!¼½

»¬-

«!!

t

t t

A A

Example 21

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Let AX = Y be a system of n linear equations in n variables.

If A ²1 exists, the solution is unique and is given by X = A ²1Y .

Pr oof

( X = A ²1Y is a solution. )Substitute X = A ²1Y into the matrix equation.

AX = A( A ²1Y ) = ( AA ²1 )Y = I n Y = Y.

( The solution is unique. )

Let X 1 be any solution, thus AX 1 = Y . Multiplying both sides of

this equation by A ²1 gives

A ²1 AX 1= A ²1Y

I n X 1 = A ²1Y

X 1 = A ²1Y.

Systems of Linear Equations

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Example 22

Solve the system 1 2 3

1 2 3

1 2 3

2 1

2 3 5 3

3 5 2

x x x

x x x

x x x

! !

! Solution

This system can be written in the following matrix form:

¼¼½

»¬¬-

«

!

¼¼

½

»¬¬

-

«¼¼½

»¬¬-

«

231

531532 211

3

2

1

x

x x

1

1

2

3

1 1 2 1 0 1 1 1 1

2 3 5 3 5 3 1 3 2

1 3 5 2 3 2 1 2 1

x

x

x

« » « » « » « » « » « »

¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼! ! ! ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ - ½ - ½ - ½ - ½ - ½- ½

The coeficient matrix is invertible and its inverse has already been

found in Example 19. We get

.1,2,1 321 !!! xxx iss lutiu iquee

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Elementary Matrices

An elementar y matrix is one that can be obtained from the

identity matrix I n through a single elementary row operation.

Example 23

¼¼

¼

½

»

¬¬

¬

-

«

!

100

010

001

3 I

1

1 0 00 0 1

0 1 0

E « »¬ ¼! ¬ ¼¬ ¼- ½

R2 m R3

2

1 0 0

0 5 00 0 1

E

« »¬ ¼!¬ ¼¬ ¼- ½

5R2

3

1 0 0

2 1 0

0 0 1

E

« »¬ ¼! ¬ ¼¬ ¼- ½

R2+ 2R1

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¼¼¼

½

»

¬¬¬

-

«!

ih g

f ed cba

A

1

1 0 0

0 0 1

0 1 0

a b c

g h i

d e

A A

f

« » « »¬ ¼ ¬ ¼

! !¬ ¼ ¬ ¼¬ ¼ ¬ ¼- ½ - ½

R2 m R3

2

1 0 0

0 5 0

0 0 1

5 5 5

a b c

d A Ae E f

g h i

« » « »¬ ¼ ¬ ¼! !¬ ¼ ¬ ¼¬ ¼ ¬ ¼- ½ - ½

5R2

32 2

1 0 0

2 1 0

0 1

2

0

a b c

d a e b f c

h i

A

g

E A

« » « »¬ ¼ ¬ ¼! !¬ ¼ ¬ ¼¬ ¼ ¬ ¼- ½ - ½

R2+ 2R1

Elementary Matrices

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E very elementary matrix is invertible. Example 24

If A and B are row equivalent matrices and A is invertible,

then B is invertible.

Pr oof

If A } « } B, then

B= En « E2 E1 A for some elementary matrices En , « , E2 and E1.

So B-1 = ( En « E2 E1 A )-1 = A-1 E1-1 E2

-1 « En -1.

1221

E I R R

}

1

1 2 0

0 1 00 0 1

E

« »¬ ¼

! ¬ ¼¬ ¼- ½

1 0 0

0 1 00 0 1

I

« »¬ ¼

! ¬ ¼¬ ¼- ½

2

1 2 0

0 1 00 0 1

E

« »¬ ¼

! ¬¬

¼¼- ½

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